ACM/ICPC 2018亚洲区预选赛北京赛站网络赛 80 Days（双向队列+尺取法）

#1831 : 80 Days

时间限制:1000ms

单点时限:1000ms

内存限制:256MB

描述

80 Days is an interesting game based on Jules Verne's science fiction "Around the World in Eighty Days". In this game, you have to manage the limited money and time.

Now we simplified the game as below:

There are n cities on a circle around the world which are numbered from 1 to n by their order on the circle. When you reach the city i at the first time, you will get ai dollars (ai can even be negative), and if you want to go to the next city on the circle, you should pay bi dollars. At the beginning you have c dollars.

The goal of this game is to choose a city as start point, then go along the circle and visit all the city once, and finally return to the start point. During the trip, the money you have must be no less than zero.

Here comes a question: to complete the trip, which city will you choose to be the start city?

If there are multiple answers, please output the one with the smallest number.

输入

The first line of the input is an integer T (T ≤ 100), the number of test cases.

For each test case, the first line contains two integers n and c (1 ≤ n ≤ 106, 0 ≤ c ≤ 109). The second line contains n integers a1, …, an (-109 ≤ ai ≤ 109), and the third line contains n integers b1, …, bn (0 ≤ bi ≤ 109).

It's guaranteed that the sum of n of all test cases is less than 106

输出

For each test case, output the start city you should choose.

提示

For test case 1, both city 2 and 3 could be chosen as start point, 2 has smaller number. But if you start at city 1, you can't go anywhere.

For test case 2, start from which city seems doesn't matter, you just don't have enough money to complete a trip.

样例输入

样例输出

2
-1

直接模拟肯定会TLE，有尺取法，可以减少很多的枚举

#include<bits/stdc++.h>
using namespace std;#define e exp(1)
#define pi acos(-1)
#define mod 998244353
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b){return b?gcd(b,a%b):a;}const int maxn=2e6+5;
int n;
ll c,a[maxn],b[maxn],s[maxn];
deque<int> q;
int main()
{int T;scanf("%d",&T);while(T--){scanf("%d%lld",&n,&c);for(int i=1; i<=n; i++)scanf("%lld",&a[i]);for(int i=1; i<=n; i++)scanf("%lld",&b[i]);for(int i=n+1; i<=2*n; i++)a[i]=a[i-n],b[i]=b[i-n];while(q.size())q.pop_back();int flag=0;for(int i=1; i<=2*n; i++){if(c+a[i]-b[i]>=0){c+=a[i]-b[i];q.push_back(i);if(q.size()>=n){flag=1;printf("%d\n",q.front());break;}}else {while(c+a[i]-b[i]<0&&q.size()){c-=a[q.front()]-b[q.front()];q.pop_front();}if(c+a[i]-b[i]>=0){c+=a[i]-b[i];q.push_back(i);if(q.size()>=n){flag=1;printf("%d\n",q.front());break;}}}}if(flag==0)puts("-1");}return 0;
}

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ACM/ICPC 2018亚洲区预选赛北京赛站网络赛 80 Days（双向队列+尺取法）

#1831 : 80 Days

描述

输入

输出

提示

ACM/ICPC 2018亚洲区预选赛北京赛站网络赛 80 Days（双向队列+尺取法）相关推荐

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